Lab goal/question: How much power/work/impulse does it take to do a double axel? What work is done with the head, hips, and feet on the take-off? Is there a significant difference?
Procedure: 1. Take a video of a jump from the internet or videotape one yourself 2. Upload the video unto LoggerPro 3. Set the scale by taking the height of the skater in meters 4. Make points by following the feet of the skater frame by frame, a graph should come up 5. Change the graph to x and y velocity 6. Find the acceleration by using the slope (on y-velocity graph) on the take-off (the steepest incline) 7. Take the weight of the skater in pounds (plus skates) and convert to kg 8. Find the upward force by taking that mass and multiplying it by the acceleration you found 9. Find the distance that the force is exerted by using the LoggerPro measurer and find work by multiplying that distance by the force 10. Find the time by looking at the graph (subtracting initial time from final time of the incline) and solve for power by taking the work divided by that time 11. Find the impulse by multiplying the force and time together 12. Go back to step 4 but this time making points by following the hips and then the head of the skater
Data/calculations:
Feet: y-acceleration on take-off: 32.23 m/s/s 110 lbs = 49.9 kg Upward force = m*a = 49.9*32.23 = 1608.28 N Upward Work = force*distance = 1608.28*0.220 = 353.82 J time = .666-.533 = .133 seconds Upward Power = work/time = 352.82/.133 = 2652.782 Watts Impulse = f*t = 1608.28*.133 = 213.9 kg*m/s
Hips: y-acceleration on take-off: 16.24 m/s/s 110 lbs = 49.9 kg Upward force = m*a = 49.9*16.24 = 810.376 N Upward Work = force*distance = 810.376*0.234 = 189.628 J time = .566-.433 = .133 seconds Upward Power = work/time = 1425.774 Watts Impulse = f*t = 810.38*.133 = 107.78 kg*m/s
Head:
y-acceleration on take-off: 26.17 m/s/s 110 lbs = 49.9 kg Upward force = m*a = 49.9*26.17 = 1305.88 N Upward Work = force*distance = 1305.88*0.226 = 295.13 J time = .6-.5 = .1 seconds Upward Power = work/time = 295.13/.1 = 2951.3 Watts Impulse = f*t = 1305.88N*.1 seconds = 130.59 kg*m/s
Reflection/Conclusion: The first thing we noticed was that the three graphs look very alike. The y-velocity goes up then goes down and then goes back to zero. But the times at which the skater accelerated were different at each body part, and so was the actual acceleration, force, work, power, impulse, etc. The greatest everything (acceleration, force, work) occured at the feet, except for the greatest power, which occured at the head. This is because the skater's knees were bent before the jump and while going up, the knees straightened out, forcing the head to go up in a shorter amount of time. (The head accelerated in .1 seconds while the feet accelerated at .133 seconds.) This makes sense because most of the jump comes from the legs, so that's why the greatest acceleration was at the feet. Also, the x-velocity changes throughout the jump in all the graphs because of the rotation, so it's not like a normal projectile where the x-velocity is the same throughout. The rotation threw us off a little because it wasn't like a normal projectile. As you can see, the acceleration from each graph was the greatest at different times. We don't know if this was maybe because of the rotation or because the hips starting rising first with the head and then the feet were last. Because of this experiment, we now know how much work and power is involved with skating. If we take the average power [2951.3 Watts + 1425.8 Watts + 2652.8 Watts)/3] which is 2343.3 Watts, and since a 60 Watt lightbulb consumes 1140 Watts per day (60x24), a simple double axel could light a 60 Watt lightbulb for over 2 days (2343.3/1140>2). That's a lot of power.
Title: Double Axel Physics Project
Lab goal/question: How much power/work/impulse does it take to do a double axel?
What work is done with the head, hips, and feet on the take-off? Is there a significant difference?
Procedure:
1. Take a video of a jump from the internet or videotape one yourself
2. Upload the video unto LoggerPro
3. Set the scale by taking the height of the skater in meters
4. Make points by following the feet of the skater frame by frame, a graph should come up
5. Change the graph to x and y velocity
6. Find the acceleration by using the slope (on y-velocity graph) on the take-off (the steepest incline)
7. Take the weight of the skater in pounds (plus skates) and convert to kg
8. Find the upward force by taking that mass and multiplying it by the acceleration you found
9. Find the distance that the force is exerted by using the LoggerPro measurer and find work by multiplying that distance by the force
10. Find the time by looking at the graph (subtracting initial time from final time of the incline) and solve for power by taking the work divided by that time
11. Find the impulse by multiplying the force and time together
12. Go back to step 4 but this time making points by following the hips and then the head of the skater
Data/calculations:
Feet:
y-acceleration on take-off: 32.23 m/s/s
110 lbs = 49.9 kg
Upward force = m*a = 49.9*32.23 = 1608.28 N
Upward Work = force*distance = 1608.28*0.220 = 353.82 J
time = .666-.533 = .133 seconds
Upward Power = work/time = 352.82/.133 = 2652.782 Watts
Impulse = f*t = 1608.28*.133 = 213.9 kg*m/s
Hips:
y-acceleration on take-off: 16.24 m/s/s
110 lbs = 49.9 kg
Upward force = m*a = 49.9*16.24 = 810.376 N
Upward Work = force*distance = 810.376*0.234 = 189.628 J
time = .566-.433 = .133 seconds
Upward Power = work/time = 1425.774 Watts
Impulse = f*t = 810.38*.133 = 107.78 kg*m/s
Head:
y-acceleration on take-off: 26.17 m/s/s
110 lbs = 49.9 kg
Upward force = m*a = 49.9*26.17 = 1305.88 N
Upward Work = force*distance = 1305.88*0.226 = 295.13 J
time = .6-.5 = .1 seconds
Upward Power = work/time = 295.13/.1 = 2951.3 Watts
Impulse = f*t = 1305.88N*.1 seconds = 130.59 kg*m/s
Reflection/Conclusion:
The first thing we noticed was that the three graphs look very alike. The y-velocity goes up then goes down and then goes back to zero. But the times at which the skater accelerated were different at each body part, and so was the actual acceleration, force, work, power, impulse, etc. The greatest everything (acceleration, force, work) occured at the feet, except for the greatest power, which occured at the head. This is because the skater's knees were bent before the jump and while going up, the knees straightened out, forcing the head to go up in a shorter amount of time. (The head accelerated in .1 seconds while the feet accelerated at .133 seconds.) This makes sense because most of the jump comes from the legs, so that's why the greatest acceleration was at the feet. Also, the x-velocity changes throughout the jump in all the graphs because of the rotation, so it's not like a normal projectile where the x-velocity is the same throughout. The rotation threw us off a little because it wasn't like a normal projectile. As you can see, the acceleration from each graph was the greatest at different times. We don't know if this was maybe because of the rotation or because the hips starting rising first with the head and then the feet were last. Because of this experiment, we now know how much work and power is involved with skating. If we take the average power [2951.3 Watts + 1425.8 Watts + 2652.8 Watts)/3] which is 2343.3 Watts, and since a 60 Watt lightbulb consumes 1140 Watts per day (60x24), a simple double axel could light a 60 Watt lightbulb for over 2 days (2343.3/1140>2). That's a lot of power.